Some hints for the evaluation of this infinite product would be appreciated.
$a_0 = 5/2$.
$a_k = (a_{k-1})^{2} - 2$ for $k\geq1.$
Evaluate : $$\prod_{k=0}^{\infty}{\left(1-1/a_k\right)}.$$
I tried finding $a_1$ $a_2$ and so on, and then multiplying them to see if I could find an easier sequence to solve but I could no proceed any further, any help would be appreciated.
Here is a kind of simplification of the problem, not an answer.
Nevertheless, the kind of methodology that is involved (introducing a geometric progression) is good to be known because it can be used in several cases.
If we write the recurrence relationship under the form:
$$\dfrac{a_k}{2}=2\left(\dfrac{a_{k-1}}{2}\right)^2-1$$
and we place it in comparison with relationship:
$$\tag{1}\cosh(2t)=2\cosh(t)^2-1$$
we can set $a_0=2\cosh(b_0)$, $a_1=2\cosh(2 b_0)$, $a_2=2\cosh(4b_0)$...
with $b_k=b_02^k$ , and $b_0=(\cosh)^{-1}(5/4)$
(a geometric sequence) we have:
$$a_k=2\cosh(b_k)$$
the infinite product becomes:
$$\tag{2}\prod_{k=0}^{\infty}{\left(1-\dfrac{1}{2 \cosh(b_k)}\right)}$$
I am rather confident that a progress has been made, and that the limit $3/7$ that we have guessed can be obtained by working on the new expression (2), for example by connecting it to the known result:
$$\prod_{k=1}^{\infty}\cosh\left(\dfrac{z}{2^k}\right)=\dfrac{\sinh(z)}{z}$$
(http://functions.wolfram.com/ElementaryFunctions/Cosh/introductions/Cosh/05/ShowAll.html)
Remark: Relationship (1) has a mirror formula $\cos(2t)=2\cos(t)^2-1$ in circular trigonometry. We haven't use it because all the $a_k$s are $>1$.
Edit: I have had a closer look at the first values of $a_n$:
$$a_0=\dfrac{5}{2}, \ \ a_1=\dfrac{17}{4}, \ \ a_2=\dfrac{257}{16}, \ \ a_3=\dfrac{65537}{256}, \cdots$$
Something has striken me: the numerators and the denominators involve particular powers of two ("powers of two that are themselves powers of 2").
More precisely, we can conjecture, and it is readily verified, that:
$$a_{n-1}=\dfrac{2^{(2^{n})}+1}{2^{(2^{n-1})}}$$
I haven't gone beyond that, but it seems a good track for further investigations.
$a_0 = 5/2 = (2^2+1)/2$
$a_1 = a_0^2 - 2 = 17/4 = (2^4+1)/2^2 $
Let $a_n = (X^4+1)/X^2, X = 2^{2^(n-1)} $
$$\begin{eqnarray} \prod_{k=0}^\infty (1-1/a_k) & = & \lim_{n\to \infty} \prod_{k=0}^n \frac{a_k-1}{a_k}\\ & = & \lim_{n\to \infty} \frac{a_0-1}{a_0} \frac{a_1-1}{a_1} \frac{a_2-1}{a_2} \cdots \frac{a_n-1}{a_n} \\ & = & \lim_{x\to \infty} \frac{2^2-2+1}{2^2+1} \frac{4^2-4+1}{4^2+1} \frac{16^2-16+1}{16^2+1} \cdots \frac{X^2-X+1}{X^2+1} \\ & = & \lim_{x\to \infty} (2^2-1)/2^2+2+1 (X^4+X^2+1)/X^4-1 \\ & = & 3/7 \end{eqnarray}$$